LintCode-A+B Problem

For given numbers a and b in function aplusb, return the sum of them.

Note

You don't need to parse the input and output. Just calculate and return.

Example

If a=1 and b=2 return 3

Challenge

Can you do it with out + operation?

Clarification

Are a and b both 32-bit integers?

    - Yes.

Analysis:

Use bit manipulation.

Solution 1:

Caluclate every bit.

class Solution {
    /*
     * param a: The first integer
     * param b: The second integer
     * return: The sum of a and b
     */
    public int aplusb(int a, int b) {
        int res = 0;
        int carry = 0;
        for (int i=0;i<32;i++){
            int a1 = a & 1;
            int b1 = b & 1;
            int val = 0;
            if (a1==1 && b1==1 && carry==1){
                val = 1;
                carry = 1;
            } else if ( (a1==1 && b1==1) || (a1==1 && carry==1) || (b1==1 && carry==1) ) {
                val = 0;
                carry = 1;
            } else if (a1==1 || b1==1 || carry==1) {
                val = 1;
                carry = 0;
            } else {
                val = 0;
                carry = 0;
            }
            val = val << i;
            res = res | val;
            a = a >> 1;
            b = b >> 1;
        }

        return res;
        
    }
};

Solution 2:

For a + b in any base, we can treat the plus as two part: 1. a + b without carry; 2. the carry generated by a +b. The a+b then equals to part 1 plus part 2. If part1+part2 generates more carry, we can then repeat this procedure, until there is no carry.

class Solution {
    /*
     * param a: The first integer
     * param b: The second integer
     * return: The sum of a and b
     */
    public int aplusb(int a, int b) {
        while (b!=0){
            int carry = a & b; //their carry (actuall, need to move to right by one bit.
            a = a^b;           //their plus result without carry.
            b = carry << 1;
        }
        return a;
    }
};